Question

(i) x, y and z are three real numbers. If x <y and z<0 then,
(a) xxz < y z (b) xxz>yxz (c) xx zs yxz (d) xx z = yxz​

Answer 1

Given : x, y and z are three real numbers. If x <y and z<0

To find : Correct option a) xxz < y z (b) xxz>yxz (c) xx zs yxz (d) xx z = yxz​

Solution:

x, y and z are three real numbers. If x <y and z<0

xxz < y z

z < 0

=> xx > y

=> x ² > y

x = 2    y  =  8

Then 2² = 4 < 8

Not valid in all conditions

xx z = yxz​

cancelling xz

=> x = y

but x < y

xxz>yxz

Dividing both sides by z

=> xx  <  yx

Dividing both sides by x

=> x < y

True if  x > 0

xxz>yxz

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Answer 2

Answer:

the correct option is option (a)